Question Number 45364 by Tawa1 last updated on 12/Oct/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}}\:+\:…\:+\:\frac{\mathrm{1}}{\left(\mathrm{2n}\:−\:\mathrm{1}\right)\left(\mathrm{2n}\:+\:\mathrm{1}\right)}\:\:=\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
![T_n =(1/((2n+1)(2n−1))) T_n =(1/2)[(((2n+1)−(2n−1))/((2n+1)(2n−1)))] T_n =(1/2)[(1/(2n−1))−(1/(2n+1))] T_1 =(1/2)[(1/1)−(1/3)] T_2 =(1/2)[(1/3)−(1/5)] T_3 =(1/2)[(1/5)−(1/7)] .... ... add them S_n =(1/2)[1−(1/(2n+1))]](https://www.tinkutara.com/question/Q45370.png)
$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\right] \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right] \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right] \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right] \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}\right] \\ $$$$…. \\ $$$$… \\ $$$${add}\:{them} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right] \\ $$
Commented by Tawa1 last updated on 12/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$